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September 10th, 2018, 01:32 AM   #1
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Dynamics 2nd question

A ship is travelling at 24 km/h in the direction N60E. At noon it is 26 km due west of a ship which is travelling south at 24km/h. Find the nearest approach of the two ships and show that this occurs at approximately 12.21 p.m. If, at noon, the captain of P had decided to intercept Q, in what direction would he have had to steer to intercept Q in the minimum time while his speed of 24 km/h.
Please... I have no idea on this topic. I just want to write solution from expert. Diagram is highly neeeded. Thanks a lot....

Last edited by skipjack; September 10th, 2018 at 04:04 AM.
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September 10th, 2018, 04:13 AM   #2
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I assume that P is the first ship mentioned and Q is the second.

You wrote "while his speed of 24 km/h", which is unclear. Were you intending it to mean "while both ships maintain a speed of 24 km/h"?
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September 10th, 2018, 04:32 AM   #3
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yes
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September 10th, 2018, 04:33 AM   #4
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Help me sir.
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September 10th, 2018, 04:51 AM   #5
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I mean P
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September 10th, 2018, 06:06 AM   #6
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I'm not getting 12.21 p.m. as the time of nearest approach, but a few minutes earlier instead. Are you sure the figures you gave are correct?
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September 10th, 2018, 08:29 AM   #7
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Please just give me the solution.
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September 10th, 2018, 09:36 AM   #8
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Quote:
Originally Posted by Harmeed View Post
Please just give me the solution.
You seem to have misunderstood what the purpose of this website is.

It is NOT to just do students homework for them.

It would be better for everyone if you just found a website more in line with just feeding you answers. They exist. This just doesn't happen to be one.
Thanks from topsquark
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September 10th, 2018, 12:36 PM   #9
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Quote:
Originally Posted by Harmeed View Post
I have no idea on this topic.
Why were you given the problem then?
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September 11th, 2018, 01:47 AM   #10
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Let's assume each ship has speed u, and that at noon P is at (0, 0) and Q is at (26, 0) (where the distance units are km).

Because P's direction is N60E, it's easy to calculate its coordinates at time t after noon. If the direction is changed to some arbitrary angle, trigonometry is needed to accomplish that.

A time t after noon, ship P has coordinates (ut√3/2, ut/2) and Q has coordinates (26 km, -ut). The distance between the ships is minimized when the square of that distance is minimized.

That squared distance, by Pythagoras, is given by (ut√3/2 - 26 km)² + (ut/2 - (-ut))²,
which equals 3(ut - (26/6)√3 km)² + (676 - 9(26/6)²) km²
which is minimized when ut = (26/6)√3 km, i.e. when t = (26/6)√3 km / (24 km/h) = 0.3127 h = 18 minutes and 46 seconds approximately.

The separation distance is then √(676 - 9(26/6)²) km = 13√3 km = 22.52 km approximately.

That answers the first part of the question. If you hand in work based on this answer, you should say that you were helped on this forum and link to this thread.
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